The generalization to the 3-D problem given is straightforward.
2.
Here x′,y′, and z′ are fixed points in space. Normally when given a charge density, we replace x,y,z with x′,y′,z′. If we did that here, we would have ρ(r′)=Qδ(0)δ(0)δ(0), which does not depend on primed coordinates. To work around this issue, change the integration variables to be double primes. As before, in two dimensions,
In the integral above, the x′ is a constant with respect to integration (in the same way that in ∫δ(x−a)dx, a is a constant) and so we replace x′′ with this constant value.
To finish the problem, I replaced ax,ay,az with the variables given in the problem of x′,y′,z′ to get the same answer as when integration was done over double primed variables.
Suppose I=∫−22g(x)dx. Find ∫−11g(2x)dx and ∫−11g(−2x)dx in terms of I. Show graphically why the result makes sense
using g(x)=x2.
Show that ∫δ(bx)f(x)dx=∣b∣f(0). (That is, show that δ(bx)=δ(x)/∣b∣).
Solution
1.
∫−11g(2x)dx=∫−11g(−2x)dx=I/2
2.
The generalization of the previous result is if I=∫−abg(x)dx, then for α>0
∫−a/αb/αg(αx)dx=∫−a/αb/αg(−αx)dx=αI.
In the case that a=b=∞ and a finite integral, it follows that ∫g(bx)dx=∫∣b∣g(x)dx, which applies to g(x)=δ(x) and thus whe have shown δ(bx)=δ(x)/∣b∣.
Griffiths 4th Edition example 1.15 gives an alternative solution. Another approach is to think of this as a transformation of variables problem in which the Jacobian, which involves the absolute value, is computed.
Find the electric field E due to a non-conducting sphere with a uniform volume charge density ρo and radius b that is centered on the origin.
Show by direct calculation that
∮SE⋅da=∫V∇⋅Ed3x
when the volume of integration V is a sphere centered on the origin with
radius b/2
radius 2b
Solution
As mentioned in class, this is somewhat of a simple problem but was given to motivate the use of the Dirac delta function. What I said in class is briefly summarized at the end.
1. The electric field is
E(r≤b)=3ϵoρorr^=4πϵoqb3rr^
E(r≥b)=3ϵoρor2b3r^=4πϵoqr21r^
where q=ρo4πb3/3 was used.
2.
In the following,
A.∇⋅(f(r)r^)=r21∂r∂(r2f(r))
is used. This follows from the identity for the divergence in spherical coordinates.
also as expected from Gauss’ law in differential form (∇⋅E=ρ/ϵo with ρ=0). Important: the last equality in the above equation is only true when r=0; when r=0, the result is 0/0. This issue is considered later in this solution.
One must split the volume integral into two integrals, one for r≤b and the other for r>b to account for the divergence being different in the two regions. As shown earlier, in the region r>b, the divergence is zero.
does not have an indeterminacy. Therefore, the divergence theorem
∫V∇⋅E∗pd3x=∮∗SEp⋅n^da
applied to E for a point charge gives
indeterminate integral=ϵoq
In this problem, it has been shown that if a point charge q is modeled as having a uniform density for r≤b and having zero density for r>b, then the indeterminacy problem can be avoided and
∫V∇⋅Ed3x=∮SE⋅n^da
for all r and for arbitrarily small (but non-zero) b (but see also ‘other notes’ below). That is, the divergence theorem gives the correct answer for a solid ball of charge that is arbitrarily small but not zero.
Although the divergence theorem does not apply for Ep, to avoid having to think about a solid ball with arbitrarily small but non zero radius is to use it with the convention that
∇⋅Ep=4πϵoq∇⋅r2r^=ϵoqδ(r)
where the function δ(r) is zero everywhere except at r=0 and integrates to 1 when the volume of integration includes r=0.
When you see the delta function used in an integration, you should think back to this problem and realize that the delta function is being used instead of modeling the electric field as a point charge q as having a uniform density for r<b and zero density for r>b.
Griffiths takes a different approach to motivating the introduction of the delta function. He notes that the indeterminate integral of the divergence
∫V∇⋅Epd3x=indeterminate integral
must be equal to q/ϵo because of the divergence theorem (which technically does not apply because of the singularity in E at r=0):
∮SEp⋅n^da=ϵoq⇒∫V∇⋅Epd3x=ϵoq
He then notes that r^/r2 has the property that its divergence
∇⋅r2r^
is zero everywhere except at the origin (where it is interminate)
integrates to a constant (4π) over any volume V that includes r=0 (based on the assumption that the divergence theorem applies and resolves the indeterminacy)
That is, although
∫V∇⋅Epd3x
has an indeterminate integrand at r=0, the divergence theorem must be true, and so this integral must be equal to the surface integral, which is ϵoq. Therefore,
∇⋅Ep=ϵoqδ(r)
where the function δ(r) is defined to be zero unless r=0 and integrates over a volume that includes r=0 to 1. This explanation is somewhat awkward because it asserts that the divergence theorem must be correct even though the field Ep does not satisfy the requirements for the divergence theorem to apply. (The requirements for the divergence theorem to apply are that the components of the vector field and their derivatives are continuous.)
In summary, the integral
∫Vf(r)∇⋅Ed3x
is indeterminant when E∼r^/r2 and V includes r=0. If we model a point charge as having a uniform density for r≤b, and zero density for r>b, where b=0 is smaller than any length scale in the problem, the indeterminacy is removed. Instead of two integrations, one for r≤b and the other for r>b, we can simply replace ∇⋅Ep with (q/ϵo)δ(r); the result of any integration will be the same.
A second motivation for using the delta function is that we often want to do an integral of the form
∫Vf(r)∇⋅Epd3x
(Recall that f(r) means f(x,y,z)). As before, integral is indeterminate because ∇⋅(r^/r2) is 0/0 at r=0. To work around this problem, we can again model the point charge as having a uniform density for r≤b and zero density for r>b. Then
The same argument applies to the higher-order terms in the Taylor series expansion.
Prior to the introduction of the delta function by Dirac in the 1930s, the above argument was used when an integral of the form
∫Vf(r)∇⋅Ed3x
was encountered and E was that for a point charge and V that included the location of the point charge.
With the introduction of the Dirac delta, one can simply use ∇⋅E=ϵoqδ(r) to get the same result without needing to think about the above steps:
∫Vf(r)∇⋅Ed3x=∫Vf(r)ϵoqδ(r)d3x={f(0)ϵoq0if V includes r=0otherwise
That is, using the delta function for the divergence of E due to a point charge will give us the same answer as if we had modeled the point charge as having a uniform density for r≤b, zero density for r>b, Taylor series expanded f(r), and then let b→0.
Other notes
In the above, I showed that the divergence theorem gave the correct result for E and any r by direct calculation. However, the divergence theorem
∮SE⋅n^da=∫V∇⋅Ed3x
requires that within V, the components of E and their derivatives are continuous. In E, Er is continuous, but ∂Er/∂r is not. So it appears that direct calculation for r>b gave a result that was consistent with the divergence theorem even though the assumptions required to apply the divergence theorem are not satisfied. To see why this occurred, use the divergence theorem in the two parts of V where the divergence theorem applies for E.
First, consider the application of the divergence theorem for the volume between r=0 and r=b.
∫Vr≤b∇⋅Ed3x=∫Sr=bE⋅n^da
The outward normal to Sr=b for Vr≤b is +r^, so
I.∫Vr≤b∇⋅Ed3x=∫Sr=bE⋅r^da
Next, consider the application of the divergence theorem for the volume between r=b and r=2b (the result will apply to any r>b). In this case, there is an inner and outer surface and so the surface integral has two parts.
∫Vb≤r≤2b∇⋅Ed3x=∮Sr=bE⋅n^da+∮Sr=2bE⋅n^da
The outward normal to Sr=b for this volume is −r^ and the outward normal to Sr=2b is +r^, so
II.∫Vb≤r≤2b∇⋅Ed3x=−∮Sr=bE⋅r^da+∮Sr=2bE⋅r^da
Using I. and II. with the volume integral expressed in two parts
∫Vr≤2b∇⋅Ed3x=∫Vr≤b∇⋅Ed3x+∫Vb≤r≤2b∇⋅Ed3x
gives
∫Vr≤2b∇⋅Ed3x=∫Sr=2bE⋅n^da
(The surface integrals at r=b cancel because their normal directions are in opposite directions.)
You do not need to turn these problems in. If you turn in these problems, I’ll provide feedback.
These are problems that either I discussed in class or are related to the topic covered in class. I place them here for your reference. Some of the problems listed may not have been covered in class but may serve as exam practice problems.
A straight 1-D rod of mass M with a uniform mass density and length a/n (n=1,2,…) is aligned with the x–axis and centered on x=a. Find the center of mass by evaluating
The step (or Heavyside step) function, Θ, is defined by
Θ(x)={10x>0x≤0
Θ can be used to to express a charge density in a compact mathematical form. For example, instead of stating “a uniformly charged sphere centered on the origin, with total charge Q, and radius b”, we can write ρ(x,y,z)=(4/3)πb3QΘ(b−r). When this density is integrated over all space, the result is Q:
Because Θ(b−r)=0 for r≥b and Θ(b−r)=1 for r<b, the limits on the r integral can be modfied and the Θ function removed:
∫ρ(x)d3x=(4/3)πb3Q∫02πdϕ∫0πsinθdθ∫0br2dr
The integrals above evaluate to the volume of a sphere of radius b so that
∫ρ(x)d3x=Q
Write the piecewise function f(x)={10∣x∣<b∣x∣≥b using one or more Θ functions.
Find the volume charge density ρ in cylindrical coordinates in terms of δ and/or Θ for an infinitely long cylinder of radius b with a charge density per unit length of λo uniformly distributed on its surface. Assume that the cylinder’s centerline is along the z-axis.
Repeat 2. assuming the cylinder is finite and extends from z=−h/2 to z=h/2.
Solution
As mentioned in class, an easy way to solve this is to write f(x) as the sum of two step functions. One shifted to the left by b, corresponding to Θ(x+b); the other shifted to the right and inverted, corresponding to Θ(x−b). Thus,
f(x)=Θ(x+b)−Θ(x−b).
Other solutions:
f(x)=Θ(b−∣x∣)
f(x)=Θ(b2−x2)=Θ[(b−x)(b+x)]
f(x)=Θ(b+x)−Θ(x−b)
It may be interesting to use dΘ/dx=δ(x) develop an identity for Θ(f(x)) similar to identity 5. on page 26 of Jackson 4th Edition.
Checks:
when x=0, f(x)=Θ(b)−Θ(−b)=1−0=1
when x=2b, f(x)=Θ(3b)−Θ(b)=1−1=0
when x=−2b, f(x)=Θ(−b)−Θ(−3b)=0−0=0
We want a charge distribution that is localized at a cylindrical radial distance b. ρ(s)=Cδ(s−b) has this feature, where C is a constant. We also need the integral of this charge density over all space to equal the total charge on the cylinder, which is Q=λoh assuming that the length of the cylinder is h. Thus, we need to find C such that
Q=λoh=∫ρ(x)d3x
Q=λoh=∫−h/2h/2∫02π∫0∞Cδ(s−b)sdsdϕdz
Q=λoh=2πhC∫0∞δ(s−b)sds=2πhCb⇒C=λo/2πb
and so
ρ(x)=2πbλoδ(s−b)
Equivalently, we can write
ρ(x)=2πsλoδ(s−b)
because δ(s−b)/s=δ(s−b)/b, which follows from the fact that the delta function is only non-zero at s=b.
Here we apply the result from part 1. with the replacement of b in that problem with h/2:
For discrete charges, Green’s Reciprocity Theorem for N charges is
i=1∑NΦiqi′=i=1∑NΦi′qi
Consider charges q1,q2, and q3 at locations x1,x2, and x3, respectively, on the x-axis. At these locations, the potentials are Φ1,Φ2, and Φ3, respectively.
If a new system of charges q1′,q2′, and q3′ is created by placing them at the same locations x1,x2, and x3, respectively, the potentials at these locations are Φ1′,Φ2′, and Φ3′, respectively.
Show that
i=1∑3Φiqi′=i=1∑3Φi′qi
Solution
The physical interpretation of Green’s Reciprocity theorem is that the work required to put charges an unprimed set of charges at x1,x2,x3 under the influence of only the electric field of a primed set of charges x1,x2,x3 must be the same as the work required to put a primed set of charges at x1,x2,x3 under the influence of only the electric field of an unprimed set of charges at x1,x2,x3.
Using
Φi=j=1j=i∑N∣xj−xi∣qj=j=1j=i∑Ndjiqj
gives, for N=3, the left-hand side terms in ∑i=13Φiqi′=∑i=13Φi′qi
In the above, the left– and right–hand sides of the equation to prove were written in the form of a matrix. In this form, one can see that the matrices are transposes of each other. If we sum all elements in a matrix, the result is the same if we sum all elements of that matrix transposed.
One can think of qj′qi/dij as elements in a matrix. The equations
i=1∑Nj=1j=i∑Ndijqj′qi and j=1∑Ni=1i=j∑Ndijqj′qi
both result in the sum of all elements of a matrix excluding the diagonals. The only difference is the order in which the elements of the matrix are summed. For a continuous charge distribution, we would use the fact that the order of integration could be changed to arrive at the result ∫dqΦ′(x)d3x=∫dq′Φ(x′)d3x′.
In class, I partially did problem 1.13 of Jackson 3rd Edition (I only found the net charge on the upper of the plate). Find the net charge on the lower plate. Justify your steps at the level of detail given in class. (I stated an easy way of finding the net charge induced on the lower plate, but I want you to do it the long way, which requires steps similar to the ones used in class.)
A point charge at a distance r from the origin is between two grounded spherical conducting shells of radius b and c that are centered on the origin. Find the net charge induced on the surfaces at b and c.
For discussion during the next class: In my solution to part 1., I used the continous form of reciprocity. Could I have solved it using the discrete form?
Solution
1.
Let the unprimed system have a conducting plate in z=0 plane held at Φ=0 and a conducting plate in the z=d plane held at at Φ=Vo.
Let the unprimed system have a conducting plate in the z=0 plane held at at Φ′=0, a conducting plate in the z=d plane held at Φ′=0, and a point charge q′ at (x,y,z)=(xo,0,0).
One can use the equation from problem 1.11 of Jackson or start with
∫allspaceρ′Φd3x=∫allspaceρΦ′d3x
and split the charge density into surface and volume charge densities σ and ρv, where the subscript v means charge not on the conducting surfaces S:
The first term on RHS is zero because ρv=0 (there are no charges in the unprimed system between the plates). The second term on the RHS is zero becuase Φ′=0 on conducting surfaces.
The first term on the LHS reduces to q′Φ(xo) becuase ρv′=q′δ(x−xo)δ(y)δ(z) (the notation ρv′=q′δ(x−xo′) is also acceptable; this gives Φ(xo), which is the same things as Φ(xo) notationally).
The potential between the plates in the unprimed system is
Φ(x)=Vodx
and so the first term on the LHS is q′Voxo/d.
The second term on LHS is composed of integrals over the top and bottom surfaces (assuming all of the charges in the primed system are on the plates or at xo; this is discussed below). Given that Φ=0 on bottom surface, we are left with an integral over upper surface (u) for which Φ=Vo.
One can reverse the potentials on the unprimed system to arrive at the charge ql′ on the lower primed surface, which is
ql′=q′(dxo−1)
Note also that the assumption that the total charge in the universe is zero
qu′+ql′+q=0
could have also been used to find ql′ given qu′.
Note: In the solution, I assumed all free charges were on the upper and lower surfaces and there were no charges on far-away surfaces; this statement is justified post hoc by uniqueness and because computing qu′ and ql′ in this way gave qu′+ql′+q=0. To avoid this assumption, one would need to use equation 1.35 of Jackson with the volume being the volume between the plates. Curiously, in problem 1.11, Jackson suggests using equation 1.35 to come up with what he calls Green’s reciprocation theorem, seemingly so it could be applied in problem 1.12. However, the equation in problem 1.11 does not directly apply because the surface of the volume between the plates is not all conductor (only the top and bottom parts of the volume are conductors, the sides of this volume are not). One is left with having to use equation 1.35 and the argument that because the problem is 1-dimensional, the derivatives of the potential with respect to the normal on the side surfaces of the volume are zero.
2.
The procedure here is nearly identical to part 1. The potential between the sphere in the unprimed system when the potential at r=b is Vo and the potential at r=c is zero is
Φ(r)=Voc−bb(rc−1)
which can be found by solving the boundary value problem
Φ(c)=0Φ(b)=Vo
∇2Φ=r21dr2d2(r2Φ)=0⇒Φ=A+B/r
to find A and B or by assuming a charge Q on the outer sphere and −Q on the inner sphere and using Gauss’ law to find Er and then V(c)−V(b)=−∫bcErdr (the Q will cancel). The final result is
In one of the steps required to derive his identities (discussed in 1.8 of Jackson, 3rd Edition), Green said: “Consider the divergence of a scalar function f multiplied by a vector function A”.
Show that
∇⋅(fA)=f∇⋅A+A⋅(∇f)
Show that
∫Vf∇⋅Ad3x=−∫VA⋅(∇f)d3x+∮SfA⋅n^da
where V is the volume enclosed by the surface S, n^ is the normal to a differential area element da on S, and d3x is a differential volume element.
Verify the result in 2. by using an f, A and V of your choosing.
Solution
1. This can be shown by writing ∇ and A in cartesian coordinates.
2. The divergence theorem
∮SU⋅da=∫V∇⋅Ud3x
using U=fA on the LHS and U=f(∇⋅A)+A⋅(∇f) the the RHS gives
∮SfA⋅da=∫V[∇⋅f(∇⋅A)+A⋅(∇f)]d3x
and rearrangement gives
∫Vf(∇⋅A)d3x=−∫VA⋅(∇f)d3x+∮SfA⋅n^da
Note that when A=Ax(x)x^, f=f(x), and V is a cube, this reduces to the integration by parts formula.
If the volume is all space, the area of S becomes infinite. If all of the charges that make up ρ and ρ′ are confined to be in a sphere centered at the origin with a finite radius, then as the radius of S→∞, E→r^/r2 and Φ→1/r and similar for E′ and Φ′. (You needed to provide some sort of justification similar to this in you solution - saying the potential and field approaches zero is not sufficient because the surface becomes infinite and you are left with ∞⋅0). So for S being a spherical surface with large r
∮SΦE′⋅n^da≃∫02π∫0πr3QQ′r2sinθdθdϕ=4πrQQ′
Where Q and Q′ are the total charges in the unprimed and primed system respectively. As a result of this argument, the surface integrals can be dropped and V can be replaced with allspace and we are left with
∫allspaceΦ′(∇⋅E)d3x=∫allspaceΦ(∇⋅E′)d3x
Finally, using ∇⋅E=−∇2Φ=ρ/ϵo and ∇⋅E′=−∇2Φ′=ρ′/ϵo gives
You do not need to turn these problems in. If you turn in these problems, I’ll provide feedback.
These are problems that either I discussed in class or are related to the topic covered in class. I place them here for your reference. Some of the problems listed may not have been covered in class but may serve as exam practice problems.
A point charge is at a distance r from the center of a conducting sphere of radius b<r. Use reciprocity to find the potential on the surface of the sphere.
Find the volume charge density ρ in cylindrical coordinates in terms of δ and/or Θ for a uniformly charged disk of radius b with charge density σo that lies in the x−y plane and is centered on the origin. Use s for the radial coordinate in cylindrical coordinates.
Use identity 5. on page 26 of Jackson to convert this to spherical coordinates. Use r for the radial coordinate in spherical coordinates.
Two infinite and grounded conducting sheets are in the x=0 and x=w plane. In the x=x′ plane, there is an infinite non-conducting sheet with surface charge density σ′.
Find the potential, ψl(x), on the left (0≤x≤x′) and to the right (x′≤x≤w), ψr(x), of the non-conducting sheet using any method (Gauss’s law or the boundary value method can be used; you should be able to do it using both methods, but you need to only show your work using one method).
Write the potential ψ(x) for 0≤x≤w as a single function using ψl and ψr and the Heavyside step function Θ. (In the future this ψ (with σ′/ϵo set to 1), will be called a Green function, which is the motivation for the title of this problem.)
Show that ∇2ψ(x)=−ϵoσ′δ(x−x′). You will need to use the fact that dΘ(x)/dx=δ(x), and dΘ(−x)/dx=−δ(x). Also compute ∇′2ψ(x), where the prime means to take derivatives with respect to primed variables. (This may seem odd because x′ was defined to be a constant; here you are being asked to treat it as a variable. You should get an answer that is proportional to δ(x−x′)).
As discussed in class, the motivation for solving this problem is that its potential, ψ, can be used in Green’s second identity (eqation 1.35), which is a form of reciprocity, to solve the most general problem for this geometry. The most general problem is to find the potential Φ(x) when ρ=ρ(x) between x=a and x=b, the left plane is grounded, and the right plane is at potential Vo. Instead of using ψ found above to solve the most general problem, first use it to solve an easier problem:
Use Equation 1.35 and ψ(x) to find the potential Φ(x) when ρ(x)=0 between the conductors and Φ(0)=0 and Φ(w)=Vo. Include a sketch or a sentence where you define V and S when you use Equation 1.35. There will be a subtilty with notation here – if you use Equation 1.35 as written, you’ll end up with Φ(x′) and not the desired Φ(x).
Solution
1. One can answer this question by assuming a surface charge of σl and σr is induced on the plates at x=0 and x=w. Then use the formula σ/2ϵo for the electric field of the sheets of charge at σl at x=0, σ′ at x=x′, and σr at x=w to get the total field El and Er. The two unknowns, σl and σr, can be found by using the fact that the electric field in either of the conductors is zero - this yields σ′=−(σl+σr), which is expected because the net charge in the universe is zero. The second equation needed to find σl and σr in terms of σ′ is
ψ(w)−ψ(0)=0=−∫0wEdx=−∫0x′Eldx−∫x′wErdx
Alternatively, one can use the fact that ∇2ψ(x)=0 has solutions of the form ψ=A+Bx, the two boundary conditions and the continuity and jump conditions at x=x′.
ψl=Al+Blxψr=Ar+Brx
Boundary conditions:
ψl(0)=0⇒Al=0
ψr(w)=0=Ar+Brw⇒Ar=−Brw
This leaves
ψl=Blx and ψr=Br(x−w)
Continuity:
ψl(x′)=ψr(x′)⇒Blx′=Br(x′−w)
Jump:
The jump condition follows from Gauss’s law with a Gaussian cylinder with one end cap to the left of x′ and the other to the right. It also follows from integrating ∇2ψ=d(dψ/dx)/dx=−σ′δ(x−x′)/ϵo once with respect to x.
The condition is
Er−El=ϵoσ′
or
[−∂x∂ψr+∂x∂ψl]x=x′=ϵoσ′
−Br+Bl=ϵoσ′⇒Bl=Br+ϵoσ′
The three equations left are
Ar=−BrwBlx′=Br(x′−w)Bl=Br+ϵoσ′
Solving gives
Bl=−ϵoσ′(wx′−1) and Br=−ϵoσ′wx′
So the final solution is
ψl=ϵoσ′(1−wx′)xψr=ϵoσ′(1−wx)x′
Note that swapping x and x′ in ψl gives the equation for ψr, and vice-versa. That is, ψl(x′,x)=ψr(x,x′).
If one plots ψ(x), you will see that its derivative has a jump downwards at x=x′ of σ′/ϵo and so we expect the terms in dψ/dx with the derivative of the Heavyside step function, which is the delta function, to be zero. So the two terms involving derivatives of Θ will be addressed first.
The first term can be rewritten using
dxdΘ(x′−x)=−δ(x′−x)=−δ(x−x′)
The first equality follows from the identity given in the problem statement: dΘ(−x)/dx=−δ(x). The second equality is a standard delta function identity.
The second term is
dxdΘ(x−x′)=δ(x−x′)
The following two equalities follow from the fact that δ(x−x′) is zero except at x=x′:
ψl(x)δ(x−x′)=ψl(x′)δ(x−x′)
ψr(x)δ(x−x′)=ψr(x′)δ(x−x′)
Given that ψr(x′)=ψl(x′), the two terms in dψ/dx involving derivatives of Θ cancel, leaving
dxdψ=Θ(x′−x)dxdψl(x)+Θ(x−x′)dxdψr(x)
Straightforward calculation and use of the same identities as above gives
dx2d2ψ=−ϵoσ′δ(x−x′)
This matches expectations. For a sheet of charge in the x=x′ plane, the volume charge density is
ρ=σ′δ(x−x′)
and Poisson’s equation is ∇2ψ=−ϵoρ
(Recall that the units of δ(x) are inverse length. This follows from part of the definition of the delta function: ∫δ(x)dx=1).
4. Equation 1.35 with ϕ replaced with Φ is
∫V(Φ∇2ψ−ψ∇2Φ)d3x=∮S(Φ∂n∂ψ−ψ∂n∂Φ)da
Let V be the volume spanned by x=[0,w], y=[0,b], and z=[0,b], where b is arbitrary. This volume has six surfaces and so the surface integral will have six parts.
In the volume integral, the term with ∇2Φ=0 because there is no charge in the volume of the Φ system. Subtitution of ∇2ψ=−(σ′/ϵo)δ(x−x′) gives
(σ′/ϵo)∫VΦδ(x−x′)d3x=∮S(Φ∂n∂ψ−ψ∂n∂Φ)da
Integration over x gives
−ϵoσ′Φ(x′)∫dydz=∮S(Φ∂n∂ψ−ψ∂n∂Φ)da
Of the six sides of the surface, Φ=0 on all except the one in the x=w plane for which the outward normal is n=x. So
where ψr=ϵoσ′(1−wx)x′ found in part 1. was used.
One could assert that we expect Φ to depend only on x, so ∂n∂Φ=0 on the other four faces, for which n=−y,y,−z,z. Alternatively, one can use symmetry make the same conclusion:
The sides of V that are in the x=0 and x=w plane have ψ=0, so two of the surface integrals of ∮ψ∂n∂Φda are zero.
On the surface in the y=0 plane, n=−y, so we need to evaluate
−∫0w∫0bΨ(x,0,z)∂z∂Φ∣∣y=0dxdz
On the surface in the y=b plane, n=y, so we need to evaluate
∫0w∫0bψ(x,b,z)∂z∂Φ∣∣y=bdxdz
Because the geometry is invariant with respect to y, the two partial derivatives are equal and the sum of these two surface integrals is zero. An idential argument can be made for the surfaces at z=0 and z=b.
All of the parts of the volume and surface integrals have not been evaluated and so equation 1.35 reduces to
−ϵoσ′Φ(x′)∫dydz=−Voϵoσ′wx′∫dydz
or
Φ(x′)=Vowx′
In the problem statement Φ(x) was requested. In the above equation, we have Φ as a function of x′, which was a constant in the ψ problem. Given that both x′0 and w, we can find any value of Φ in this range by solving a problem with x′ set to that value. Thus, we can equivantly replace x′ with x and so we have shown
Two conducting and grounded spherical shells of radius b and c are centered on the origin, and c>b.
A nonconducting spherical shell is centered on the origin and has a charge density of σ′ and radius r′, with b<r′<c.
Find ψi(r), the potential between the inner conducting shell and the charged shell and ψo(r), the potential between the charged shell and the outer conducting shell. (Read the subscript i as “inner” and o as “outer”.)
Find the surface charge densities on the inner and outer conductor.
Verify that Gauss’s law is satisfied for a Gaussian sphere centered on the origin and with a radius that is between the charged shell and the outer conducting shell.
Write the potential ψ(r) for b≤r≤c as a single function using ψi and ψo and the Heavyside step function Θ.
Solution
1.
Instead of starting with ψi=Ai+Bi/r and ψo=Ao+Bo/r, we immediately write
ψi=Ci(r1−b1) and ψo=Co(r1−c1)
because we know the potential must have a 1/r dependence and the above two equations satisfy ψi(b)=0 and ψo(c)=0. What is left then is to use the conditions ψi(r′)=ψo(r′) and Eo(r′)−Ei(r′)=σ′/ϵo to find the two unknows Ci and Co. The result is
ψi=−ϵoσ′r′2(b1−c11)(r′1−c1)(r1−b1)
ψo=−ϵoσ′r′2(b1−c11)(r′1−b1)(r1−c1)
Written in terms of the total charge q′ on the shell with charge density σ′ at r=r′, this is
ψi=−4πϵoq′(b1−c11)(r′1−c1)(r1−b1)
ψo=−4πϵoq′(b1−c11)(r′1−b1)(r1−c1)
As before, there is symmetry with respect to swapping the spatial coordinate r with a parameter r′: ψi(r,r′)=ψo(r′,r) (note that this is only after writing the potentials in terms of the charge on the shell at r=r′).
2. From Gauss’s law, near the surface of a conductor σ=−ϵodψ/dn, where n is in the direction perpendicular to the conductor and outwards. For the inner conductor, n=r; for the outer, n=−r. (Note that in Green’s second identity, the convention is that n is in the direction perpendicular to V and outwards.) Evaluation givves
σi=−4πq′(b1−c11)(r′1−c1)[b21]
σo=+4πq′(b1−c11)(r′1−b1)[c21]
The charge induced on the inner and outer surface is
qi=−q′(b1−c11)(r′1−c1)
qo=+q′(b1−c11)(r′1−b1)
Checks:
1. As r′→b, qi→−q′ and qo→0.
2. As r′→c, qi→0 and qo→−q′.
3. In HW #2.3.2, we found the charge induced when a point charge was at r′. We can use that answer and superposition to find the total charge induced if point charges are distributed uniformly on a sphere. This should match the total charge induced found in this problem. From #2.3.2, the answers were (with ro replaced with r′):
qr=b′=−c−bq′b(r′c−1)=−q′(b1−c11)(r′1−c1)
qr=c′=+c−bq′c(rob−1)=+q′(b1−c11)(r′1−b1)
The “superposition” argument requires a bit more justification. Consider two separate problems. In one problem there is dq′ at (r′,θ′,ϕ′); call the potential for this single dq problem ψ′(r,θ,ϕ;r′,θ′,ϕ′). This potential will satisfy
∇2ψ′(r,θ,ϕ;r′,θ′,ϕ′)=−dq′δ(r−r′) with ψ′=0 at r=b and ψ′=0 at r=c
In another problem, there is dq′′ at (r′,θ′′,ϕ′′); call the potential for this single dq′′ problem ψ(r,θ,ϕ;r′′,θ′′,ϕ′′). This potential will satisy
∇2ψ′′(r,θ,ϕ;r′′,θ′′,ϕ′′)=−dq′δ(r−r′′) with ψ′′=0 at r=b and ψ′′=0 at r=c
The sum ψ′+ψ′′ satisfies
∇2(ψ′+ψ′′)=−dq′δ(r−r′)−dq′δ(r−r′′) with ψ′+ψ′′=0 at r=b and ψ′+ψ′′=0 at r=c
This argument can be extended for an arbitrary number of differential charges on a sphere of radius r′.
We can conclude that the sum of the potentials for isolated charges at any position on a surface of radius r′ and r′′ satsifies Poisson’s equation and the boundary conditions, which is the same condictions that we require for the combined charge problem. The charge density obeys superposition: σ/epsilono=−dψ/dn=−(dψ′/dn+dψ′′/dn) so that the charge densities on the conductors for two charge case is the sum of the charge densities for from each individual case.
Another way of justifying superposition is to note that for the isolated dq′ problem, the net force on the charges induced on the conductors is zero. If dq′′ is introduced the net force on all of the induced charges will still be zero.
Before stating Equation 1.42, Jackson (3rd edition) notes that with Equation 1.35 and 1.39, it is simple to obtain a generalization of Equation 1.36.
Show the “simple” steps need to arrive at Equation 1.42.
Be very careful with notation. You’ll need to think a bit about the justification and validity of swapping x and x′ and/or changing the dummy variable used in integration. When you do either of these, provide a justification so that I know you are not applying the “answer operator”.
Solution
In HW 3.1, the use of Green’s second identity resulted in a potential that depended on the primed variable because integration was performed over an unprimed area or volume. In that problem, an argument was given for why the prime variable could be simply replaced with the unprimed variable. In principle, one could always use the approach used in HW 3.1 to find Φ(x′) and then make an argument for why Φ(x) is the equation for Φ(x′) with the prime removed. In this problem, we follow Jackson’s logic for Equation 1.42 for which the result of integration is Φ(x), which depends on the unprimed variable.
We want the result after integration to depend on x, so change the integration variable to be primed. The derivatives must also be changed to be primed and G(x,x′) must be replaced with G(x′,x).
can be used to evaluate the first term in the volume integral, where the prime means the partial derivatives are taken with respect to primed coordinates. The first term evaluates to
For a general function, say, g(x,y)=xy2, g(x,y)=xy2=g(y,x)=yx2. Showing G(x,x′)=G(x′,x) in general is non-trivial and Jackson problem 1.14 gives a suggestion for the proof, and a footnote on page 40 gives a reference of the proof for 1.14(b). So ideally one would have cited the statement on page 40 of Jackson to justify the use of G(x,x′)=G(x′,x).
In class, I showed how to find the potential ψ due to a point charge inside of a grounded, conducting, and closed dome when the radius of the dome is infinite. The potential was found using the method of images. (A dome is a hemisphere plus a cap.)
Use this ψ and equation 1.35 of Jackson (or equation 1.42; ψ is proportional to a Green function) to find the potential Φ for the following problem:
An conducting dome has its base in the x−y plane and the radius of the dome is infinite. There are no charges inside of the dome. The boundary of the dome is grounded except for the region between x2+y2<b, which is held at a potential of Vo.
Find the potential Φ(z) inside of the dome (the integral required to find Φ(x,y,z) requires an approximation of the integrand or numerical integration and will be considered in the future).
Find the potential Φ(z) inside of the dome when b/z≪1. Do this by Taylor series expanding Φ(z) for small b/z. Does the answer make sense?
Sketch the electric field lines inside of the dome.
Note that a related problem (but with a different Green function) considered in 2.6 of Jackson.
Solution
1. The potential that satisfies the boundary conditions in the limit that that that the radius of the dome is much larger than r′ is
This equation was derived using the method of images in class and is the same equation that solves the problem for a point charge at x′ above an infinite grounded plane in the x-y plane. ψ satisfies the boundary conditions that when z=0, ψ(x,y,0)=0 and when r′≫r, ψ=0 (showing the latter requires re–writing the denominators in terms of r′ and r).
Given that we are only interested in the potential along the z–axis, we can set x′=y′=0. (In 2.6 of Jackson, he does this after the integral required to find Φ(x,y,z) is found.) We then have
ψ=x2+y2+(z−z′)2kq−x2+y2+(z+z′)2kq
Green’s second identity (Equation 1.35) with Φ in place of ϕ is
∫V(Φ∇2ψ−ψ∇2Φ)d3x=∮S(Φ∂n∂ψ−ψ∂n∂Φ)da
In this problem V is the volume of the dome and S has two surfaces – a large disk in the x–y plane and the curved part of the dome.
The Laplacian of ψ can be written down by inspection. The charge density is that for point charges at (x,y,z)=(0,y,±z′), so
Inside V, we were given that there are no charges, so ∇2Φ=0. Therefore, the left-hand side of Equation 1.35 reduces to −(q/ϵo)Φ(z′).
On S, ψ=0, so the second surface integral is zero. We are left with
−ϵoqΦ(z′)=∮SΦ∂n∂ψda
The surface is composed of the two parts – the curved part and its base. On the curved part, Φ=0 was given. On the base, Φ=0 for x2+y2<b2 and Φ=Vo otherwise. As a result, the surface integral reduces to
−ϵoqΦ(z′)=∫x2+y2<b2Vo∂n∂ψda.
In cylindrical coordinates, we have
−ϵoqΦ(z′)=Vo∫02π∫0b∂n∂ψsdsdϕ
The outward normal to V is -z, so
−ϵoqΦ(z′)=−Vo∫02π∫0b∂z∂ψsdsdϕ
What remains is to evaluate the partial derivative (and recall that ∂ψ/∂z in the integral means ∂ψ/∂z∣on S). In this problem, “on S” corresponds to z=0.)
∂z∂ψ∣∣z=0=2kqs2+z′23z′
Evaluation of the integral gives
−ϵoqΦ(z′)=−4πkqVo[∣z′∣z′−b2+z′2z′]
Cancelling constants and using z′/∣z′∣=1 for z′>0 gives
Φ(z′)=Vo[1−b2+z′2z′]
Using the same arguments used in HW 3.1, we can replace z′ with z, leaving
For large z, the (positive) charge on the disk appears as a point charge at the origin. As a result, we may expect that the potential will fall off as 1/z. Here the potential falls off more slowly. The reason is that there is also a contribution from the negative charges in the plane for s>b. As a result, for large z, the system appears to have zero charge at the origin and so there is no 1/z term. (The interpretation of the power series expansion of potential will be covered in more detail later.)
3. This was discussed in class. In general, it is easiest to draw equipotential lines starting with one near the boundary. Then, draw electric field lines that are perpendicular to the equipotential lines.
Review Griffiths 3.3.1 (3rd and 4th edition). You should be able to quickly derive the potential for Figure 3.20 (3rd and 4th edition) when any one of the sides is at Vo and the other three are grounded. See also my PHYS 305 notes.
Suppose the sides of Figure 3.20 of Griffiths 3rd or 4th edition are held at Vl, Vr, Vt, and Vb, where l=left, r=right, t=top, and b=bottom. Find the potential inside the rectangular pipe.
Hints:
The total potential will be the the sum of the potentials from four different problems (1)
Vl non–zero and all other sides zero, (2) Vr non–zero and all other sides zero, etc. I will ask you to explain why in class.
You can build the solution to three of the four problems described in the first hint using a transformation of coordinate system and the solution to one of the problems.
Solution
Note: Diagram and solution below assumes width is b and height is a. Diagram in Griffiths has width of 2b and height of 2a.
The potential inside of the long (in z direction) duct with cross-section is shown in the following figure
The choice of coordinate system for I. was arbitrary and we can translate the duct used for I. to the left by b/2 and down by a/2 so that the position of the duct is the same as that in Figure 3.20 of Griffiths.
This is effected by replacing x with x+b/2 and y with y+a/2 in I.:
In 2.7 of Jackson 3rd edition, he solves for the potential outside of a sphere with hemispheres held at ±V using a Green function. In example 3.7 of Griffiths 4th edition (example 3.6 in 3rd edition), he describes how to solve for the potential outside of a sphere with a potential of V(θ) on its surface. As a result, Jackson’s example is related to Griffiths’ example.
Study these two examples and be prepared to answer questions (such as how they are related) about them in class.
In Example 3.3 of Griffiths, he finds the potential for a long duct with one open face. The diagram for that example is similar to the above except there is no conductor on the right side and b=∞. The solution he finds is
II.V(x,y)=π4Von=1,3,…∑∞ne−nπx/asin(nπy/a)
1. Show that the infinite b solution (II.) matches the finite b solution (I.) in the limit that a/b≪1. (You do not need to derive I. or II., but should be able to.)
2. In the limit that b/a≪1, it would seem that near the center of the duct (y=a/2), the system appears as two infinite parallel conducting plates held at different potentials, which is a 1-D problem solved before. Show that in the limit b/a≪1, equation I. reduces to the solution for an infinite conducting plate in the x=0 plane held at Vo and a grounded infinite conducting plate in the x=b plane. (You’ll need to use eu≃1+u for u≪1 multiple times and also the Gregory and Leibniz formula for π).
Solution
1. We need to show that
sinh(nπb/a)sinh[nπ(b−x)/a] reduces to e−nπx/a when a/b≪1, or, equivalently, b/a≫1
Re-writing the numerator and denominator in terms of exponentials
An infinitely long, hollow, and conducting duct is parallel to the z-axis and has sides bounded by 0≤x≤1 and 0≤y≤1. All sides are grounded. An infinitely long non-conducting sheet of charge with surface charge density σ′ is in the x=x′ plane between y=0 and y=1. See the left part of the figure below.
1. Find equations ψl(x,y), the potential for x<x′, and ψr(x,y), the potential for x>x′. Based on the discussion in class, we can easily write down equations for ψl and ψr that are zero on the conducting walls:
ψl=∑n=0∞Ansin(nπy)sinh(nπx)
ψr=∑n=0∞Bnsin(nπy)sinh(nπ(x−1))
where n is an integer, matches the boundary conditions on the walls of the conductor. You need to find An and Bn such that the two conditions
ψl(x′,y)=ψr(x′,y)
and
[−∂x∂ψr+∂x∂ψl]x=x′=ϵoσ′
are satisfied.
2. Write a single equation for the potential ψ(x,y) using ψl, ψr, and the Heavyside step function Θ.
The Green function is G=(4πϵo/Aσ′)ψ, where A=∫01dy∫0Ldz and L is the length of the duct. (You may want to verify this.)
3. Use ψ(x,y) and Green’s second identity (Equation 1.35 of Jackson), or equivalently, the Green function stated above and equation 1.42 of Jackson to find the potential when the duct is filled with a nonconducting material with a uniform charge density ρo as shown in the right part of the figure above.
Note: In problem 2.15 of Jackson, he gives the Green function for the case where a line of charge parallel to the z-axis passes through (x,y)=(x′,y′). That problem is a bit more complicated than the problem considered here. In problem 2.16, of Jackson, he states the result of using the Green function from problem 2.15 to find the potential for the configuration in part 3. above.
Solution
1. Both
ψl=∑n=0∞Ansin(nπy)sinh(nπx)
and
ψr=∑n=0∞Bnsin(nπy)sinh(nπ(x−1))
satisfy ∂2ψ/∂2x+∂2ψ/∂2y=0 and the boundary conditions
ψl(0,y)=0ψl(x,0)=0ψl(x,1)=0
ψr(1,y)=0ψr(x,0)=0ψr(x,1)=0
The equations for ψl and ψr could have been derived using the above six boundary equations and starting with
However, the result will be the same. In problems such as these, it is often easier to reason out what the form of the solution will be rather than starting from the most general equation. In this case, we need the sin solution in the y direction because it can have zeros at two locations. The sinh(nπ(x−1)) term is justified because we know that Asinhx and Bcoshx can be combined into a Csinh(x+δ) with a “phase” shift δ.
The interpretation of this equation is that to find the potential at x, sum over the potential due to differential sheets at different x′. The first integral is the potential just to the left of all differential sheets to the right of x. The second integral is the potential just to the right of all differential sheets to the left. This is the equation you would probably write down in order to compute the potential due to a slab if you did not know about Green functions – the total potential is simply the sum of potentials due to all differential sheets.
A non-conducting and infinitely long cylindrical shell of radius b has a surface charge density of σ=σ2sin2ϕ.
Find ψ(s,ϕ) using the boundary value method. Note that the general solution to Laplace’s equation in cylindrical coordinates for a problem with no z dependence is
Repeat 1. assuming σ=σo+σ2sin2ϕ. In this case, the lns term cannot be dropped for the s>b solution; ao is determined from the requirement that for large s, the potential should be that for a line of charge with linear density λ that is the average of σ over ϕ.
Solution
I’ll solve part 2., which reduces to part 1. if σo=0. I will also solve this the long way. After solving this problem the long way, you should be able to identify short–cuts, one of which is find the potnential due to a long and uniformly charged cylinder and adding the result to the answer of part 1.
We can conclude that the sn terms are zero for the potential outside of the cylinder because at large s the cylinder will be approximately that for a line of charge, for which potential will be proportional to λlns for large s, where λ is the azimuthally averaged surface charge density: λ=∫02πσbdϕ. That is, for large s, the potential should be the same as that when all of the charge on the cylinder is collapsed onto a line, for which the potential is ψ=const−(λ/2πϵo)lns, giving ao=−λ/2πϵo and Ao=const. The constant Ao is determined by defining a reference potential Vr at a reference point sr. For example, if ψ(sr)=Vr, then Vr=Ao−λlnsr/2πϵo and the terms Ao+aolns combine to form
Ao+aolns=Vr−(λ/2πϵo)ln(s/sr)
Note For a line of charge, we can’t choose a reference potential to be zero at infinity. This is related to the problem that arises if one attempts to use V=4πϵo1∫dq/∣r−r′∣ to compute the potential for an infinite line of charge. The derivation of this integral assumes that the potential is zero at large r. Using Gauss’ law, one can show that this is not the case – Es=2πϵoλs1 and so ψ=ψo−2πϵoλlns, where ψo is a constant. We can’t choose ψ(s=∞)=0 because it gives ψo=−∞. So to compute the potential for an infinite line of charge, one must either use the integral equation for the electric field or Gauss’ law. See also 2.3.4 of Griffiths where he notes that the integral diverges for a charge distribution that extends to infinity.
Outside the cylinder, the general form is then
ψo=Ao′+aolns+∑n=1∞(ancosnϕ+bnsinnϕ)s−n
I’ve kept the lns term. In the following, I’ll show how the values of Ao′ and ao given above follow from an alternative argument than the one given above.
Inside the cylinder, the potential should be finite and so the lns and s−n terms are dropped. (If the cylinder had a constant charge density, we would expect the potential to be constant inside of the cylinder and would keep only the Ao term.)
Inside the cylinder, the general form is then
ψi=Ao+n=1∑∞(Ancosnϕ+Bnsinnϕ)sn
The two boundary conditions are
ψo(b,ϕ)=ψi(b,ϕ) and [∂ψo/∂s−∂ψi/∂s]s=b=−σ/ϵo
(Make sure that you know how to derive the second boundary condition.)
from “matching terms”, which means integrating both sides by either cosn′ϕdϕ or sinn′ϕdϕ, with n′ being an integer greater than zero, and integrating from ϕ=0 to 2π and using
∫02πcosndϕ=0;∫02πsinndϕ=0;∫02πcosn′ϕsinnϕdϕ=0
∫02πcosn′ϕcosnϕdϕ=0 if n′=n and π if n′=n
∫02πsinn′ϕsinnϕdϕ=0 if n′=n and π if n′=n
Notice that in cylindrical problems we integrate over ϕ, so the limits of integration are from 0 to 2π. In contrast, in spherical problems with a θ dependence, we integrate from 0 to π.
To determine the relationship between the coefficients quoted above, it is often easier to write out the first few terms in the sums
The equation ao=−bσo/ϵo from the second boundary condition combines with Ao′+aolnb=Ao from the first boundary condition to give
Ao=Ao′−(bσo/ϵo)lnb
The equation a1/b=A1b from the first boundary condition combines with −a1/b2−A1=0 from the second boundary condition to give
2A1=0
which is only possible if A1=0. Similarily, all terms except b2 and B2 are zero.
The equation b2/b2=B2b2 from the first boundary condition and −2b2/b3−2B2b=−σo/ϵo combine to give
B2=+σ2/4bb2=b3σ2/4
The final result is
ψi=Ao′−(bσo/ϵo)lnb+4ϵ0σ2bs2sin2ϕ
ψo=Ao′−(bσo/ϵo)lns+4ϵ0σ2s2b3sin2ϕ
If the reference potential is chosen to be zero at s=0, then Ao′=(bσ1/ϵo)lnb and
ψi=4ϵ0σ2bs2sin2ϕ
ψo=(bσo/ϵo)ln(b/s)+4ϵ0σ2s2b3sin2ϕ
(In this problem, the reference potential was not given intentionally. Ideally you recognized that a reference potential was needed to fully specify the solution. Also note that we cannot specify a constant reference potential at any other s because the potential depends on ϕ.)
Note: Inspection of this answer suggests a faster way of solving the general problem of
σo+σ1sinϕ+σ2sin2ϕ+…
or more generally,
σo+σ1ssinϕ+σ1ccosϕ+σ2ssin2ϕ+σ2ccos2ϕ+…
First find the potential due to σo. Then, write ψi and ψo not as an infinite sum, but rather as a sum that involves only terms of sinnϕ and cosnϕ that appear in σ.
Check: Earlier it was argued that for large s, ψo should approach
ψo(s→∞,ϕ)→const−(λ/2πϵo)lns
where
λ=∫02πσbdϕ
Using σ=σo+σ2sin2ϕ gives λ=2πbσo and so the computed value of ψo of
Due Sunday, March 13th at 11:59 pm. Email me a scan (if it is less than 10 pages), slide under my door (preferred), or bring to the next class (if you finish early).
Do problem HW problem 5.2.1 with the modification that σ′=λ′δ(y−y′) and show that how it can be used to arrive at the result quoted in problem 2.15(b) of Jackson.
Use your answer to part 1. and reciprocity to find the potential inside the tube when ρ=0 inside the tube, the side at x=1 is held at a potential of Vo, and the other three sides are grounded. (This is an analog to HW #3.1.4 where we solved a problem that was easy to solve without reciprocity using reciprocity).
1. Find Φ(r,θ) for problem HW 4.1 in terms of an infinite sum involving the Legendre polynomials.
2. Find the surface charge density on the part of the dome in the x–y plane.
Notes
Try to work out σ using only the first few terms in the expansion. This will require only straightforward derivatives and the use of the table of Legendre polynomials.
Given that the exam is coming up, I’ll make getting the general equation extra credit. As a hint, use the middle equation of 3.29 of Jackson, which is dPl+1(u)/du=udPl(u)/du+(l+1)Pl(u) to compute the derivatives of the Legendre polynomial. Given that we are always evaluating at u=0 (corresponding to θ=π/2), you need to use dPl+1(u)/du∣u=0=(l+1)Pl(0). This will give you a general equation for the derivative evaluated at u=0 … but you’ll need to know Pl(0). You can get that using the first equation of 3.29 of Jackson: (l+1)Pl+1(u)−(2l+1)uPl(u)+lPl−1(u) with u=0. Extra extra credit if you plot the result.
In class, we started the problem of finding the potential inside and outside of a sphere of radius b that is centered on the origin and is held at potential V(b,θ)=Vocos2θ.
Find Φ(r,θ) (unless otherwise stated, this type of statement means for all r).
Most students immediately re–wrote cos2θ as Vo(Po/3+2P2/3) prior to starting to answer part 1. by using the fact that P2=(3cos2θ−1)/2 and Po=1. In the general case, it will not always be obvious how to express an arbitrary boundary potential V(θ) in terms of the Legendre polynomials.
In 3.2 of Jackson, he notes that an arbitrary function of θ (he uses x, which is also cosθ) may be written as
V(θ)=l=0∑∞AlPl(cosθ)
To find the coefficients Al, multiply both sides by Pl′(cosθ)sinθdθ and then integrate from 0 to π. (This is essentially what Griffiths calls “Fourier’s trick” except using Legendre polynomials). By orthogonality, one then finds that
Al=22l+1∫0πV(θ)Pl(cosθ)sinθdθ
Suppose V(b,θ)=Vo for θ=0 to θ=π/2 and V(b,θ)=−Vo for θ=π/2 to π. Write V(b,θ) in the form A0P0+A1P1+A2P2. That is, find a second–order approximation to V(b,θ) by finding A0, A1, and A2. (If you use a formula from Jackson or elsewhere instead of doing integration to find A0,A1, and A2, prove the formula.)
Find Φ(r,θ) using the boundary value V(b,θ)=A0P0+A1P1+A2P2.
State at least one way that you can determine if your answer to part 3. makes sense.
(Optional) Repeat this problem assuming that instead of being held at a potential, the sphere has a surface charge density of σ(b,θ)=σocos2θ or σ(b,θ)=σo for θ=0 to θ=π/2 and σ(b,θ)=−σo for θ=π/2 to π
